Hey guys I have a few problems related to a figure I'd like to understand the problem and learn how to solve it, I think Im just a little unclear on it but anyways any help is welcomed :)

- anonymous

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

##### 1 Attachment

- anonymous

Im trying it out again i think i may have just mistaken something

- anonymous

1 is choice 3: 112

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

Ok 2 3 & 4 lost me :S

- saifoo.khan

how can u say 1) 112?

- anonymous

can someone help break them down a bit plz?

- saifoo.khan

i dont think 1) = 112..

- saifoo.khan

180 - 112 = ?

- anonymous

well ACF covers the same length as GAC

- saifoo.khan

Nope, |dw:1331792427015:dw|

- saifoo.khan

the lengths dosnt really matter..

- anonymous

oh its the angels
not the lengths of the line :\

- anonymous

ps sorry open study kinda crashed google chrome

- anonymous

So how do i go about finding the angle lengths for the first problem:? so I can figure out Part one of the question?

- AccessDenied

its a parallelogram, so AG is parallel to CF with a transversal AC, and m< GAC = 112
|dw:1331792928009:dw|

- AccessDenied

soo that would make ACF the "same-side interior angle" to GAC, which means they will be supplementary. That makes sense?

- anonymous

so i was right m

- S

1.
|dw:1331793329783:dw|
Do you see how GAC and CFG look the same? Meaning they are both more then 90 degrees.. they are opposite or something i forgot what its called cause its been a while but they are opposite and equal to each other. Same way ACF and AGF are equal to each other.. they are both sharp.. both less then 90 degrees..
Now the sum of all 4 angles is 360. AGF + GAC = 180 or ACF + CFG = 180 or CFG + FGA = 180 or GAC + ACG = 180

- S

As we said:
GAC + ACF = 180
They gave us that GAC = 112, So lets plug in and solve for ACF
112 + ACF = 180
ACF = 180 - 112
ACF = 68

- anonymous

Wow that was vary clear Id like help on the other problems, but first can you test me on a another problem like this one? i want to see if i really got it

- S

Ok I ll help you with the rest of them. Yes, lets try a similar problem.
|dw:1331794002046:dw|
Angle D is equal to 73 degrees. Find:
Angle A
Angle B
Angle C

- anonymous

B has to be 73 and A & C must be 107 theya ll then add up to 360

- S

Yup! That's perfect!

- anonymous

YEA!!! :)

- S

So all of them add up to 360 or like neighboring pairs add up to 180.. like
A+B = B+C = C+D = D+A = 180

- S

Ok lets do number 2 now =)

- S

We are dealing with AGF and GAC. Just by looking at them we know that they add up to180.

- S

So if AGF + GAC = 180, lets now plug in the values that are given to us
(a+20) + (2a+10) = 180
a + 20 + 2a + 10 = 180
3a + 30 = 180
3a = 180 - 30
3a = 150
a = 50
Now we need to find GFC which is the same as GAC which they said is (2a + 10)
Plug in 50 for a
2a+10 = 2*50 + 10 = 110 degrees

- S

So our GFC, (as well as GAC) is equal to 110 degrees

- S

For number 3
|dw:1331794863730:dw|
AC and GF have to equal to each other. In a parallelogram, opposite sides are parallel and equal, so AC = GF and AG = CF

- S

AG = CF
2x + 10 = 5x - 2
10 + 2 = 5x - 2x
12 = 3x
x = 12/3
x = 4

- anonymous

Ok I get 3 but not 2 lol 2 was a mind Puck

- S

Ok question 2
Look at angles AGF and GAC. They are neighbors.. one looks like he is less then 90, other one looks to be more then 90 correct? So they add up to 180. Do you understand this part?

- anonymous

Yes I do

- S

Ok, so AGF = GAC
In the question they said that AGF = (a+20) degrees and GAC = (2a+10) degrees correct?

- anonymous

a+20 <= this has me saying wtf

- anonymous

is it just there to add to the problem?

- S

Yea I know, but just look at it as a regular number. You know like you have for example
x + 2 = 3
x is just a hidden unknown number.. and you find it by x = 3 - 2... x = 1
Same thing here.. instead of giving you a regular number, they give a hidden unknown number, which in this case is not (x), but an (x + 20), or as they put it an (a + 20)

- S

So instead of giving you AGF = 70 degrees for example,
they give you AGF = (a+20) degrees.
They want you to find a and then use it in order to find the angle

- S

And the only way we can solve for from this is to set it equal to something that we know it actually equals. In this case, sum of these two angles AGF and GAC is equal to 180

- anonymous

OK err thats still going to get me its just one of those things lol but i understand it now so thats good:)

- S

|dw:1331795961782:dw|

- S

Ok, well its just the same as we had before. One angle plus another is equal to 180.
In this case instead of having numerical values we have unknown and numerical values, but by putting them together its still equal to 180. So from there we find our unknown a:
So if AGF + GAC = 180, lets now plug in the values that are given to us
(a+20) + (2a+10) = 180
a + 20 + 2a + 10 = 180
3a + 30 = 180
3a = 180 - 30
3a = 150
a = 50
Now we need to find GFC which is the same as GAC which they said is (2a + 10)
Plug in 50 for a
2a+10 = 2*50 + 10 = 110 degrees

- S

Ok where we left off? 3 or 4? =D

- anonymous

4 And man your My hero :)

- S

Hehe, no problem man, don't mention it =D This stuff helps your brain, and exercises mine =D So I benefit from it as much as you do =D

- S

Ok now question 4. |dw:1331796416239:dw|
First look at the square ABCD. Its diagonals AC and BD bisect each other. They intersect at point O and cut each other in half. So AO = CO = BO = DO

- S

Now it works similarly with a parallelogram ACGF. AF and CG intersect each other at point X.
They bisect each other, or cut each other in half.
So, AX = XF and CX = XG.
Note that this part is not the same as the one we had with a square, where aaaallllll the little pieces AO BO CO DO were equal to each other. Here, only AX = XF and CX = XG, because if you see, in a parallelogram, diagonals are not equal, AF is shorter than CG. So once again, here AX = XF and CX = XG.

- S

Now lets get to our question. We know that
AX = XF
Now we plug in the given values for AX and XF
(4y+3) = (2y+7)
And from here we solve for y.
4y + 3 = 2y + 7
4y - 2y = 7 - 3
2y = 4
y = 2

- anonymous

First of all Id like to sum up my thoughts on this, Loads of steps lol looks complex at first sight but it is simple when someone smart breaks it down :)
I would also like to give a BIG Thank you And let you know I have learned a lot in the pas hour working with you

- anonymous

Really I cant thank you enough :) I really got it & my friend it was all thanks to you :-D

- S

Hehe, yea, it may seem confusing at first, but that's all it is about, breaking it down into small and simple parts and solving it nicely and simply =)
Noo problem man, don't worry about it, I'm glad I could help you =) That's why we are on this website in the first place =D Give help and get help =D

- anonymous

Amen to that man :)

- S

Lol =D

- S

Ok, anytime you need help with this type of stuff, geometry and stuff, let me know, send me a message or something, I ll try my best to explain it to you )

- anonymous

Again id like to let you know im walking away with a better understanding of how to better dissect problems :) And i will :) ps you should add me as a fan xD so i can see when your on

- S

Ok, glad I could help you =) Yup, good idea, just added you! =)

- S

Ok, I'm gonna take a look at a few more problems and go to sleep now =D

Looking for something else?

Not the answer you are looking for? Search for more explanations.